Ta có: \(x^2-6x-7< 0\)
\(\Leftrightarrow x^2-6x+9-16< 0\)
\(\Leftrightarrow x^2-6x+9< 16\)
\(\Leftrightarrow\left(x-3\right)^2< 16\)
\(\Leftrightarrow x-3< \pm4\)
\(\Leftrightarrow\left[{}\begin{matrix}x< 7\\x< -1\end{matrix}\right.\)
Vậy: Với x < 7 hoặc x < -1 thì x2 - 6x - 7 < 0